Electronic – using a NOR gate output to a mosfet switch, improper high level output voltage

You are looking for the Gate Threshold Voltage, marked Vgs(th) in the datasheet. There is a bit of a catch to watch out for here though.

For this particular MOSFET, if we look at the relevant bit of the datasheet:

We can see it's given as a minimum of 0.6V, and a maximum of 1.2V. This looks promising for your 1.8V logic output. However, the catch is that this is for a drain current of only 250uA, so it's not of much practical use, since we'll need more than that to pull the PMOS gate low with the 4.7kΩ resistor, we need to make sure the drain current is acceptable at 1.8V.
To get a better idea of what voltage we need at the gate to sink useful currents at the drain, we would usually need to take a look at the graphs for Id over Vgs and Id over Vds for different gate voltages is also useful:

Unfortunately both only goes as low as a gate voltage of 2.25V, so we don't really know how much current the drain will sink at 1.8V. Since at 2.25V Id is over 10A, it's a reasonably safe bet that at 1.8V we'll be fine sinking 3.3V / 4.7kΩ = 0.7mA, but we can't be certain, which is ideally what we need to be.

Solution options:

• Use a different N-ch MOSFET with a guaranteed drain current of over 0.7mA at 1.8V Vgs given in the datasheet.
• Use any small NPN transistor which has a Vbe of ~0.7V and will be turned on easily with 1.8V (with current limiting base resistor of course. For a gain of e.g. 100, assuming you want to pull as low as possible, calculate using (1.8V - 0.7V) / (0.7 / 100) = ~150kΩ, then halve this to compensate for gain reduction at saturation, e.g. 75kΩ or less. Another conservative rule of thumb is to assume a gain of 20-30)
• Do your own tests with a few IRLML2502s and confirm there is more than enough drain current at a Vgs of 1.8V.
• Increase R2, so less current is needed to pull the gate down.

Personally I'd go for using an NPN, since it's cheaper - you could get a suitable part for a couple of pence, compared to the 33 pence for the MOSFET.
For sticking with this part though, unless you need fast switching, or there is a lot of noise present, I'd probably go for simply increasing R2, to around 15kΩ, so you only have to sink 3.3V / 20kΩ = 220uA. This is less than the value given at a max of 1.2Vgs, so you can be certain it will be able to pull the gate down easily.

One other possible option which I use regularly is to use an IO in open drain mode - many microcontrollers have certain pins which are tolerant of higher voltages than their supply voltage, so if your micro has a 3.3V tolerant pin which can be used in open drain mode (you don't give the part number so I can't check this, although you may have done so already) then you can do away with Q1 completely and use this instead.

That LED being in the emitter - to - ground connection of Q3A is going to give you some grief since you've only got 3.3V (less actually because of the R7 R8 voltage divider on Q3A's base) to drive Q3A with.

Fix it by connecting Q3A emitter directly to ground, lose R8 and drive the base through R7, and connect the LED from the source of Q1 to ground through a suitable ballast resistor.

From your question:

"OUTPUT4 is the high power load output, which is connected to the Battery negative terminal through the system's common ground." might be better worded to indicate that one end of the load - not Q1's source - will be connected to battery minus.